From few days I'm trying to make this code work:
.PHP :
if (($_SERVER['REQUEST_METHOD'] === 'GET') && isset($_GET['tkn']) && isset($_GET['t'])) {
$tkn = $mysqli->real_escape_string($_GET['tkn']);
$t = $mysqli->real_escape_string($_GET['t']);
if (time() < $t) {
$info["msg"]="<strong>Error! </strong>link valid!";
$info["cls"]="brand";
}
else {
$info["msg"]="<strong>Error! </strong>Link expired!";
$info["cls"]="danger";
}
$mysqli->close();
json_encode($info);
}
.JS :
$.ajax({
url: window.location.href,
success: function(response) {
console.log(response);
},
error: function(response) {
console.log(response);
},
dataType: "JSON"
});
If I use the current PHP code without adding exit(json_encode($info)); It will return the entire html tag without any text returned by PHP code.
If I'm adding exit(json_encode($info)); it will display only that code on white background.
Any ideas? Thanks!
Here's my login.php format
<?php if ($_SERVER['REQUEST_METHOD'] === 'POST') { //some code goes here echo json_encode($info); exit; // this part is working perfectly, is getting the answer from php via ajax and display it in page(html part in running) } elseif ($_SERVER['REQUEST_METHOD'] === 'GET') { //other code goes here echo json_encode($info); exit; // here is displayed only the message(without html part) if I add exit; //and if i dont add exit is displayed the message on white background on // the top of the page(html part is working); ?> //html part
You need to print the json string after all
use
echo json_encode($info);
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